2
$\begingroup$

The book I'm following for Strength of materials states that, in a beam with symmetric cross section (about the plane of bending) and in pure bending, the moment of cross sectional area about the neutral axis is zero, i.e.

$$\int_A y\,dA=0$$

where $y$ is the perpendicular distance of an area $dA$ from the neutral axis.

The book then says that, this could only happen when the neutral axis passes through the centroid of section, and I don't understand why.

I have a feeling this might be very trivial to ask, but please bear with me.

enter image description here

Improve this question
$\endgroup$
0

3 Answers 3

Reset to default
1
$\begingroup$

The neutral axis intuitively is an axis that if you support a section on its neutral axis on a straight edge the section will balance itself. Or mathematically if we get the area moment about the neutral axis it should give us zero moments otherwise the section will rotate.

The moment of a differential element with an area dA about this axis is its distance (y)* area (dA).

$$M_{dA}=y dA$$

$$\sigma M_{neutral- axis}=0$$

$$\int ydA=0$$

For example, if we have a square obviously, the neutral axis is at its middle and divides the square into 2 rectangular, one above one below the neutral axis.

$$\sigma M=0 \rightarrow A_{upper - rect}* dx- A_{lower-rect}*dx=0$$

$$A/2 * A/4- A/2 * A/4=0$$

Improve this answer
$\endgroup$
4
  • $\begingroup$ I get that moment of the cross sectional area about the neutral axis is zero, but how does one conclude from that, that neutral axis is passing from the centroid of the c/s? Neutral axis as my textbook says is the axis where neutral surface intersects the c/s, and where nuetral surface is the surface which experiences no change in length and stress. This neutral axis could come anywhere, how did we conclude it comes right at the centroid? $\endgroup$
    – Harshit Rajput
    Feb 1, 2022 at 6:42
  • $\begingroup$ you can prove that by taking the area moment about y Axis. $ \int xdA=0 $ they intersect at Centrid. $\endgroup$
    – kamran
    Feb 1, 2022 at 6:47
  • $\begingroup$ I'm not able to follow that, could you please walk me through it, whenever you get time. $\endgroup$
    – Harshit Rajput
    Feb 1, 2022 at 6:51
  • 1
    $\begingroup$ intuitively centroid of a section is a point which if you hang it from that point the section is at equilibrium and remains balanced. That is the intersection of the horizontal and vertical neutral axis. $\endgroup$
    – kamran
    Feb 1, 2022 at 6:56
1
$\begingroup$

In any cross-section, the neutral axis is located at where the bending stress is zero. In the case of asymmetric cross-section, it falls on the center of the section (where the centroid is located) with the areas above and below the neutral axis are equal, and $y_{top} = y_{bottom}$.

enter image description here

Improve this answer
$\endgroup$
0
0
$\begingroup$

So we know that the moment of the cross sectional area about the neutral axis is zero that is,

$$\int_A y\,dA=0$$

where y is the perpendicular distance from the neutral axis of an area $dA$

Let the origin of the coordinate system lie at the center of Neutral Axis. The $y$ in the above equation is the distance measured along the y axis of this coordinate system.

enter image description here

The centroid C, of this cross sectional area will lie somewhere on the y axis. The y coordinate of the centroid is given by,

$$\bar y= \frac{\int_A y\,dA}{\int_A dA}$$

since $$\int_A y\,dA=0$$

$$\bar y=0$$

This suggests that the centroid is at $y=0$, i.e. on the Neutral axis. Conversely, the neutral axis passes through the centroid of the cross sectional area.

enter image description here

Improve this answer
$\endgroup$
2
  • $\begingroup$ This derivation tells you the location of neutral axis, not the direction of plane of N.A. The plane of N.A is always perpendicular to the direction of lateral load. $\endgroup$
    – Rameez Ul Haq
    Feb 1, 2022 at 20:01
  • $\begingroup$ My question was on the location of the neutral axis itself, that how does one conclude it passes through the centroid. $\endgroup$
    – Harshit Rajput
    Feb 2, 2022 at 9:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.